Tag Info

New answers tagged

1

One of the main tools on infinite matrices and Hilbert spaces operators is the so-called Schur's test. This is Exercise 45 in Halmos' A Hilbert Space Problem Book. Schur's test. Let $A=[a_{ij}]_{i,j\in\mathbb{N}}$ be an infinite matrix. Suppose that there exist positive numbers $p_i>0$, $q_j>0$ ($i,j\in\mathbb{N}$), $\beta>0$ and $\gamma>0$ ...


0

Yes, you are right. In the following I will try to explain how actions of $G$ and representations of $G$ are related, as I believe this was your main problem here. An action of $G$ on $W$ is a map $G \times W \rightarrow W, (g,w) \mapsto g.w$ which has the two properties $1.w = w$ and $(gh).w = g.(h.w)$ for all $g,h \in G$ and $w\in W$. Thus the maps $w ...


1

Call $Px$ the orthogonal projection of $x$ onto $M$, then it's easy to see that $\| x-Px\|= \inf_{y\in M}\| x-y\|$. On the other hand we have that $\max_{y\in M^\perp, \| y\|=1} |\langle x,y\rangle|$ is just the norm of the linear functional $l(y)=\langle x,y\rangle $ when viewed as $l\colon M^\perp \to \mathbb{R}$. Since $Px \perp M^\perp$ we have that $$ ...


1

The image of $T$ is also closed: Take $T(x_k)\rightarrow y_0 \in H_2$. Then $T(x_k)$ is Cauchy. Since $T$ is bounded below, the sequence $x_k$ is also Cauchy. So $x_k\rightarrow x_0$. So $T(x_0)=y_0$.


0

Here is the easiest way to go. Choose a sequence $\{Y_n\}$ with $Y_n\to Y$ in $\mathcal{L}^2$. You can pass to a subsequence so that $Y_n \to Y$ almost everywhere. This tells you that $Y$ is measurable in any $\sigma$-algebra for which the $Y_n$ are measurable.


3

Consider $$T:\ell^2\rightarrow\ell^2\\T(x_n;\ n\in\mathbb{N}^+)=\left(\frac{x_n}{n};\ n\in\mathbb{N}^+\right)$$ Notice that $\|T\|\leq1$, but $\left(\frac{1}{n};\ n\in\mathbb{N}^+\right)\notin T(\ell^2)$ But $\forall g\in \ell^{2*}\ (g\circ T=0\rightarrow g=0)$ because $T(\ell^2)$ contains a Hilbert basis. So the answer to your question is no.


1

Let $ \varphi: \mathcal{H} \to \mathbb{C} $ be any (continuous) linear functional on $ \mathcal{H} $. Then the composition $$ \varphi \circ T = \left\{ \begin{matrix} {\ell^{2}}(\mathbb{N}) & \to & \mathbb{C} \\ (a_{n})_{n \in \mathbb{N}} & \mapsto & \sum_{n \in \mathbb{N}} a_{n} \varphi(f_{n}) ...


1

HINT: Try to prove the following: (1) If $T$ is invertible then so is $T^*$ and $(T^*)^{-1} = (T^{-1})^*$. (2) If $T$ is invertible, then $T^{-1}$ is a bounded linear operator (Open Mapping Theorem).


1

You need to show that $(\mathrm{ran}\, A)^\perp = \{0\}$ if and only if $\ker A = \{0\}$. The key is that $A^*$ is injective on the range of $A$: if $A^*Ax = 0$ then $$ 0 = \langle x,A^*Ax \rangle = \langle Ax , Ax \rangle = \|Ax\|^2 $$ so that $Ax = 0$. Similarly, $A$ is injective on the range of $A^*$. This means that $$\ker A = \ker A^*A \quad ...


2

As KCd said in a comment, a continuous linear functional on a Hilbert space $H$ is uniquely determined by its action on a given orthonormal basis. This follows since, if $M$ is an orthonormal basis of $H$, then for any $x\in H$ we have $$f(x)=\sum_{m\in M}\langle x,m\rangle f(m)$$ which follows from continuity. If $\dim H=\infty$ and we do not require $f$ to ...


2

The fact that such a $ * $-homomorphism exists is the statement of the Gelfand-Naimark-Segal (GNS) Construction.


2

Let $ \mathcal{H} $ be a Hilbert space and $ (\mathbf{e}_{i})_{i \in I} $ a Hilbert basis of $ \mathcal{H} $. If $ f $ is a linear mapping from $ \mathcal{H} $ to $ \mathbb{C} $, then $ f $ is continuous if and only if $$ (\spadesuit) \qquad (f(\mathbf{e}_{i}))_{i \in I} \in {\ell^{2}}(I). $$ Hence, although a linear functional on $ \mathcal{H} $ is uniquely ...


4

Here is another answer that is suitable for the poster’s background. We make use of the fact that for a self-adjoint bounded operator $ T $ on a Hilbert space $ \mathcal{H} $, we have $$ \| T \|_{B(\mathcal{H})} = \sup(\{ |\langle T(h),h \rangle_{\mathcal{H}}| \mid h \in \mathbb{S}(\mathcal{H}) \}), $$ where $ \mathbb{S}(\mathcal{H}) $ denotes the unit ...


4

As $ T $ is bounded and self-adjoint with norm $ 1 $, its spectrum $ \sigma(T) $ is a compact subset of $ [-1,1] $, and its spectral radius $ r(T) $ equals $ 1 $. Hence, either $ -1 \in \sigma(T) $ or $ 1 \in \sigma(T) $. If $ -1 \in \sigma(T) $, then $ 0 \in \sigma(I + T) $, and so $ I + T $ is not invertible. If $ 1 \in \sigma(T) $, then $ 0 \in \sigma(I ...


0

If $E_N$ is the spectral resolution for $N$, then the following is bounded and normal by the functional calculus: $$ N(I+N^{\star}N)^{-1} = \int \frac{\lambda}{1+|\lambda|^{2}}dE_{N}(\lambda). $$ In line with your comment under this point that you are trying to derive the spectral measure, let's try a different approach. For any $x$, one has ...


0

Ok, let's be clear on something first: Linear algebra per se is not really important in a serious study of quantum mechanics. This is because most of the spaces one studies in quantum theory-Hilbert spaces, Heisenberg state spaces,etc-and their linear mappings,are actually infinite dimensional function spaces. Linear algebra proper is really the study of ...


2

$$\langle x+\alpha y,x+\alpha y\rangle\geq\langle x,x\rangle\forall \alpha\\ \langle x,x\rangle+2\alpha\langle x,y\rangle+\alpha^2\langle y,y\rangle\geq\langle x,x\rangle\forall\alpha\\ 2\alpha \langle x,y\rangle+\alpha^2\langle y,y\rangle \geq0\forall\alpha$$ But the left-hand side equals zero if $\alpha=0$ and if $\alpha=-2\langle x,y\rangle/\langle ...


0

Orthogonal complements are always closed, so you find $$ H = M^\perp \oplus M^{\perp\perp}. $$ Moreover, it holds $M^{\perp\perp} = \bar M$, which is the closure of $M$. To see this, first one can prove $M^\perp = (\bar M)^\perp$. This implies $$ M^{\perp\perp} = (\bar M)^{\perp\perp} = \bar M, $$ where the latter equality uses closedness of $\bar M$. Thus, ...


1

First of all you should understand that what is meant when one says that "$\textit{Not all normed linear spaces are inner product spaces.}$" Here is the explanation. An $\textit{inner product space}$ is a vector space with an inner product defined on it. An inner product on $X$ defines a norm on $X$ given by \begin{equation} \|x\| = \sqrt{\langle x,x ...


4

The space of continuous functions on $[0,1]$ with the norm $\|x\|= \sup\{|x(t)| : t\in[0,1] \}$ is a Banach space. To justify that claim, you would need to know that all continuous functions on $[0,1]$ are bounded, and that Cauchy sequences in this metric space actually converge, and those take some work. One way to prove that this norm does not come from ...


2

The completion of a normed space $X$ can be constructed as a quotient of the space of Cauchy sequences of $X$, where the quotient identifies sequences $x_n$ and $y_n$ with $\| x_n - y_n \| \to 0$. The natural norm of this quotient space, let's call it $\hat X$, is $\| [x_n] \|_{\hat X} := \lim_{n \to \infty} \| x_n \|$. Following this lead a natural ...


3

No, it's not true. If $\{ e_{n} \}_{n=1}^{\infty}$ is an orthonormal sequence in an infinite dimensional Hilbert space, then this sequence converges weakly to $0$ because $\sum_{n=1}^{\infty}|(x,e_n)|^{2} \le \|x\|^{2}$ for all $x$, which forces the general term of the sum to converge to $0$. If $\{ x_n \}$ converges weakly to $x$ and $\{ \|x_n\| \}$ ...


3

In $\mathcal{L}^2([0,2\pi])$, the functions $x \mapsto \sin(nx)$ converge weakly to 0. However their $\mathcal{L}^2$ are all the same. Your conjecture does not hold.


0

On section 4 of Von Golitschek, Manfred, Ming-Jun Lai, and Larry L. Schumaker. Error bounds for minimal energy bivariate polynomial splines. Numer. Math (1999), authors show that problem of finding a spline with minimal energy surface can formulated into a standard approximation problem in Hilbert space. For DLSQ splines also we can follow their attitude. ...


4

Contrary to what I tried to do first, I think I now have a (complete) argument that your claim is false. Let me rewrite your last assumption (the displayed equation). Since $\int x^2|\widehat{g}(x)|^2 = \int |g'|^2$, this really says that certain derivatives are uniformly bounded in $L^2$ (originally, I may have to define these as distributional ...


0

Let $\sigma_n$ list all finite sequences of $0, 1$. Fix a bijection $f:\{ \sigma_n : n \geq 1\} \to \mathbb{N}$. Let $e_n = \langle 0, 0, \dots, 1, 0, 0, \dots \rangle \in l^2$ (1 occurs at the nth position). For each $x \subseteq \mathbb{N}$, let $a_x = \sum \left(\sqrt{3/2}\right)2^{-n} e_{f(x \upharpoonright n)}$. Now check.


3

No, it is not uniform. For all $N$, $\|(1_H-T_N)e_{N+1}\|=1$, showing that $1_H-T_N$ has norm at least $1$. It is exactly $1$, as $1_H-T_N$ is the orthogonal projection onto the closed span of $\{e_n\}_{n\geq N+1}$. More generally, note that the set of compact operators is norm closed, so a noncompact operator cannot be uniformly approximated by compact ...


2

Without loss of generality $A$ acts non-degenerately on $H$, i.e. $$\cap_{X\in A}\ker X=\{0\},$$ otherwise consider the restriction of $A$ onto the orthogonal complement of $\cap_{X\in A}\ker X$. Let $\tilde A$ be the unitization of $A$. Then $\tilde A$ is naturally isomorphic to $A+\mathbb C I_H\subseteq B(H)$. The notion of spectrum of an element $X\in A$ ...


3

If this were the case, then any closed, densely defined operator on $H$ would automatically be bounded. Indeed, let $A$ be a closed, densely defined operator on $H$ with domain $D(A)$. By definition of a closed operator, the graph $$\Gamma(A):=\{(x,Ax)\in H\oplus H\,:\,x\in D(A)\}$$ is a closed subspace of $H\oplus H$. If your question had a positive answer, ...


1

No, this need not be the case. If you take $H := {\mathscr l}^2({\mathbb N})$, you can identify the projection $H\oplus H\to H$ with the map $\text{even}: {\mathscr l}^2({\mathbb N})\to {\mathscr l}^2({\mathbb N})$ mapping a sequence $(x_n)_n$ to the sequence $(x_{2n})_n$ of its even parts. Now, consider the subspace $U := \{(x_n)\in {\mathscr l}^2({\mathbb ...


0

You know that an operator is invertible if and only if it's injective and surjective. It's clear that $(\lambda-T)$ is always injective, hence the point spectrum is empty. Now try to figure out surjectivity. You should find that the spectrum is $[0,1]$. Edit : If $(\lambda-T)$ is surjective, then for each $g\in L^2([0,1])$ there should be an $f\in ...


0

Every Hilbert space is isomorphic to $L^2(X)$ for some measure space $X$. This is a generalization of Tomek's answer. Now this is redundant, since $X$ is not determined up to isomorphism by $H$. However, this formulation sheds light onto the spectral theorem: If $A$ is a self-adjoint (possibly unbounded) operator $H \rightarrow H$, then there is a measure ...


1

On the one hand, you can apply the same thing as you did with a generic vectors space and obtain that $$H \cong \bigoplus\limits_{i \in X} \mathbb{C}.$$ Notes that this relies on the notion of Hamel dimension of the space of a vector space. However, you can say that a Hilbert space has its Hilbert dimension equal to the cardinality of a certain set $E$, ...


2

Every Hilbert space is isometrically isomorphic to $\ell_2(\Gamma)$ for some set $\Gamma$. This follows directly from Parseval's identity.


2

I found this notation in Simon's paper "Hamiltonians defined as quadratic forms", Commun. Math. Phys. 21 (1971), 192-210. (PDF available on his web page). A footnote at the bottom of the first page reads: $$X + (L^\infty)_\varepsilon = \{ f \mid (\forall \varepsilon) f = x_\varepsilon + g_\varepsilon \text{ with } x_\varepsilon \in X; ...


1

Let $(g_{n})$ be a sequence in $K$ satisfying $\left\|f-g_{n}\right\|_{p}\rightarrow\delta:=\inf_{g\in K}\left\|f-g\right\|_{p}$. It follows from the triangle inequality that the $g_{n}$ are bounded in norm, and therefore lie in a closed, convex subset $E\subset K$, which is weakly compact by Alaoglu's theorem. Whence $(g_{n})$ has a subset ...


2

For $L^1$, take $K$ to be the set of functions for which $\int h \mathrm{d}x = 0$. Let $f$ be the characteristic function of the interval $[0,1]$. Note $$ 1 = \int h - f \mathrm{d}x \leq \|h - f\|_1 $$ Note that this minimum is achieved for any $h$ of the form: $h = +1$ on a measure 1/2 subset of $[0,1]$, $h = -1$ on its complement in $[0,1]$, and $h = 0$ ...


1

Hint: $$\|f\|^2 = \sum_{k=1}^n \|f^{(k)}\|_2^2 \ge \|f^{(k)}\|_2^2$$


1

One example might be $M(K)$, the space of all regular Borel measures on $K$ of finite variation, where $K$ is compact space. This space arises as the dual of $C(K)$.


1

In fact all normed spaces are subspaces of some function spaces. This could be the reason why functional analysis have its name.


1

$\mathbb{R}$ is an infinite dimensional vector spaces over $\mathbb{Q}$, but it is not an Hilbert space. You can see that all the axioms for a vector space are verified if you define the sum of two ''vectors" as the usual sum of real numbers and the product for a scalar $q \in \mathbb{Q}$ as the usual product. This space has an infinite dimensional Hamel ...


0

No, it is not possible that an infinite dimensional Hilbert space is the linear span of a countable subset. It is obvious that the whole space is the smallest closed subspace containing a dense subset. So as you indicated in a comment, it could be a typo where $\overline{\mathrm{span}}$ was intended instead of $\mathrm{span}$, to indicate that one must ...


2

By definition $$\hat u(\xi)=\int e^{-ix\xi}u(x)dx$$ We also know $\widehat{\partial^ku(\xi)}=i^{|k|}\xi^k\widehat{u(\xi)}$ Similar to the integration by part step posted in my question, we can get $$\int(\Delta u)^2=\sum_{i,j=1}^n\int(\widehat {u_{x_i x_i}} \widehat{u_{x_j x_j}})=\sum_{i,j=1}^n\int (-\xi_i \xi_i \hat u)( -\xi_j \xi_j \hat ...


1

Regarding a) In fact, the definition of span of a set is the set of finite linear combination of element of respective set, i.e. $$ {\rm span}(M):= \{\sum_{i=1}^n c_i m_i | n\in {\mathbb N}, m_1,\ldots m_n\in M \& c_1,\ldots, c_n\in {\mathbb F}, \} $$ So you can do as you have said. Regarding b) One may discuss by taking an arbitrary element $x$ in ...


1

$$||x+\lambda y||\geq ||x||\iff ||x+\lambda y||^2\geq ||x||^2\iff P(\lambda)= 2Re[\overline{\lambda}\langle x,y\rangle] +|\lambda|^2||y||^2\ge0$$ If $y=0$ the answer is trivial. Assume therefore $y\neq 0$. Then, if we select $$\lambda_0=-\frac{\langle x,y\rangle}{\|y\|^2}$$ we have $$P(\lambda_0)=-\frac{|\langle x,y\rangle|^2}{\|y\|^2}$$ which is ...


1

For a counterexample to your claim, notice that $C^\infty_c(\Omega)$ is a subspace of $H^1_0(\Omega)$ (although it is not closed), and its dual is $D'(\Omega)$ (the space of all distributions on $\Omega$) which is not a subset of $H^{-1}(\Omega)$. Note that to use the Hahn-Banach theorem, you need a sublinear function which is defined on the whole space and ...


1

$$ \begin{align} 2\Phi(x+h)-2\Phi(x) & = (T(x+h),x+h)-(Tx,x) \\ & = (Tx,x)+(Tx,h)+(Th,x)+(Th,h)-(Tx,x) \\ & = (Tx,h)+(Th,x)+(Th,h) \\ & = (Tx,h)+(h,Tx)+(Th,h) \end{align} $$ I assume you must be working on a real space, and working with an operator $T$ defined on the whole space, which makes it ...


0

Choose any $n$ such that $a < r_n < b$. Then $$|F(x)| \geq 2^{-n}|f(x-r_n)|$$ so for all $x \in (r_n, b)$ we have $$|F(x)|^2 \geq 2^{-2n}|f(x-r_n)|^2 = \frac{2^{-2n}}{x-r_n}$$ Then $$\begin{aligned} \int_{a}^{b} |F(x)|^2 dx &\geq \int_{r_n}^{b}|F(x)|^2 dx \\ &\geq 2^{-2n}\int_{r_n}^{b}|f(x-r_n)|^2 dx \\ & = ...


1

Fixed $k$, for all $j$, we have $\langle v_j,X(v_k)\rangle=\langle v_j,Y(v_k)\rangle \Leftrightarrow \langle v_j,X(v_k)\rangle - \langle v_j,Y(v_k)\rangle =0 \Leftrightarrow \langle v_j,(X-Y)(v_k)\rangle=0$. But this implies that $(X-Y)(v_k)$ is orthogonal at all $v_j$, hence $(X-Y)(v_k)=0$. That is, $X$ and $Y$ coincide in a base. Therefore, $X=Y$.


0

For operator $A^*=0$ the condition $\Vert A^*g\Vert=1$ is always false. By the fundamental rules of logic False implies anything and in particular that $\Vert g\Vert\leq 1/\delta$. One could say for example that $\Vert A^*g\Vert=1\implies\pi=10^5$ and would be absolutely correct.



Top 50 recent answers are included